Plot the flight envelope (combined V-n diagram) for the following GA acrobatic aircraft. Then, determine the maximum load factor.

m = 2300 kg, S = 19.33 m², $C_{L\max}$ = 2, − $C_{L\max}$ = −1.2, AR = 7, $C_{Lα}$ = a = 6.3 1/rad, $V_c$ = 310 KEAS (at 10,000 ft).

Question

Plot the flight envelope (combined V-n diagram) for the following GA acrobatic aircraft. Then, determine the maximum load factor.

m = 2300 kg, S = 19.33 m², [latex]C_{L\max}[/latex] = 2, −[latex]C_{L\max}[/latex] = −1.2, AR = 7, [latex]C_{Lα}[/latex] = a = 6.3 1/rad, [latex]V_c[/latex] = 310 KEAS (at 10,000 ft).

Accepted Answer

FAR does specify regulations/requirements for gust loads as well as a positive maximum load factor and a negative maximum load factor for acrobatic aircraft. We first need to calculate the aspect ratio and mean geometric chord:

[latex]AR=\frac{b^2}{S}\Rightarrow b=\sqrt{AR \cdot S}=\sqrt{7 imes 19.33}\Rightarrow b=11.63\space m \quad \quad \quad \quad (3.9) \ \space \ \overline{C}=\frac{S}{b}=\frac{19.33}{11.63}\Rightarrow \overline{C}=1.66 \space m \quad \quad \quad \quad (3.9) [/latex]

The combined V-n diagram is plotted in three steps: (1) basic (maneuver) V-n diagram, (2) gust V-n diagram, and (3) combined V-n diagram.

1. Basic V-n diagram

As discussed in Section 9.9.3, the general shape of the basic V-n diagram resembles Figure 9.30. We just need to determine the coordinates of points K, J, G, F, B, and A. The unit of velocity in the graph is selected to be in knot (KEAS).
Since the aircraft type is acrobatic, the maximum limit load factor, based on FAR 23, is as follows:
(Positive) [latex]n_{\max}[/latex] = +6,
(Negative) [latex]n_{\max}[/latex] = −(0.5 × 6) = −3.
The dive speed of this aircraft from Equation 9.200 is
[latex]V_D=1.55V_C=1.55 imes 310=480.5\space knot[/latex].
Hence, the coordinates of points F and G are (6, 480.5) and (−3, 480.5). To determine coordinates of points A, B, J, K, we need to derive two equations regarding [latex]C_{L\max}[/latex].

[latex]V_s=\sqrt{\frac{2mg}{ ho SC_{L_{\max}}}}=\sqrt{\frac{2 imes 2300 imes 9.81}{1.225 imes 19.33 imes 2}}=30.87\space m/s=60 \space KEAS \quad \quad \quad \quad (2.49)[/latex]

Thus, the coordinate of point A is (1, 60). The top stall curve or the load factor as a function of airspeed (in m/s) is

[latex]n=\frac{ ho V^2SC_{L_{\max}}}{2W}=\frac{1.225 imes V^2 imes 19.33 imes 2}{2 imes 2300 imes 9.81}=0.100105V^2\quad \quad \quad \quad (9.195)[/latex]

For point B, the load factor is given as 6, so the corresponding speed is determined by

[latex]6=0.100105\space V^2 \Rightarrow V=75.6\space m/s=147\space KEAS[/latex]

Thus, the coordinates of point B is (6, 147). With the same technique, we can derive the equation for the lower stall curve

[latex]V_{si}=\sqrt{\frac{-2mg}{ ho S(-C_{L_{\max}})}}=\sqrt{\frac{-2 imes 2300 imes 9.81}{1.225 imes 19.33 imes (-1.2)}}=39.85\space m/s=77.5 \space KEAS \quad \quad \quad \quad (2.49)[/latex]

So, the coordinate of point K is (−1, 77.5). The lower stall curve or the load factor as a function of airspeed (in m/s) is

[latex]-n =\frac{ ho V^2 S(-C_{L_{\max}})}{2W}=\frac{1.225 imes V^2 imes 19.33 imes (-1.2)}{2 imes 2300 imes 9.81}=-0.00063V^2 \quad \quad \quad \quad (9.195)[/latex]

For point J, the load factor is given as −3; hence, the corresponding speed is determined as

-3 = -0.00063 V² ⇒ V = 69 m/s = 134.2 KEAS

Thus far, we have collected the following coordinates:

O → (0, 0),
A → (1, 60),
B → (6, 147),
F → (6, 480.5),
G → (-3, 480.5),
J → (-3, 134.2),
K → (-1, 77.5).

By using these data, we can plot the basic V−n diagram as illustrated in Figure 9.30.

2. Gust V-n diagram:

The variations of load factor as functions of airspeed (V) and gust speed ([latex]V_g[/latex]) are given by

[latex]n=1+\frac{k_gV_{gE}V_Ea ho S}{2W} \quad \quad \quad \quad (9.206)[/latex]

Since the cruising speed is given for 10,000 ft, two flight conditions are considered for the maximum load factor. Then, we calculate n for both [latex]V_C and V_D[/latex].

a. Sea level
The aircraft mass ratio is

[latex]\mu_g=\frac{2m}{ ho \overline{C}as}=\frac{2 imes 2300}{1.225 imes 1.66 imes 6.3 imes 19.33}=18.75 \quad \quad \quad \quad (9.208)[/latex]

Then, the gust alleviation factor is obtained as

[latex]k_g=\frac{0.88\mu_g}{5.3+\mu_g}=\frac{0.88 imes 18.75}{5.3+18.75}=0.684\quad \quad \quad \quad (9.207)[/latex]

From Table 9.9, at sea level with the cruising speed, the gust velocity of ±50 ft/s (i.e., ±15.25 m/s) should be considered; thus, the load factor will be

[latex]n=1+\frac{0.684 imes (\pm15.25) imes V imes 6.3 imes 1.225 imes 19.33}{2 imes 2300 imes 9.81}\Rightarrow n=1\pm 0.03436V\quad \quad \quad \quad (9.206)[/latex]

Since the cruising speed ([latex]V_C[/latex]) is 310 KEAS,

n = 1 + 0.03436V = 1 + 0.03436 × 310 × 0.5144 = 6.48 (positive value)

n = 1 - 0.03436V = 1 - 0.03436 × 310 × 0.5144 = -4.48 (negative value)

We perform similarly for dive speed. When the aircraft is flying with dive speed ([latex]V_D[/latex]), the gust speed should be ±25 ft/s (i.e. ±7.5 m/s). Hence, the load factor is

[latex]n=1+\frac{0.684 imes (\pm 7.5) imes V imes 6.3 imes 1.225 imes 19.33}{2 imes 2300 imes 9.81}\Rightarrow n=1\pm 0.01688V\quad \quad \quad \quad (9.206)[/latex]

Since the dive speed ([latex]V_D[/latex]) is 480.5 KEAS,

n = 1 + 0.01688V = 1 + 0.01688 × 480.5 × 0.5144 = 1 + 4.173 = 5.173 (positive value)

n = 1 - 0.01688V = 1 - 0.01688 × 480.5 × 0.5144 = 1 - 4.173 = -3.173 (negative value)

b. 10,000 ft altitude
At 10,000 ft altitude, the air density is 0.9 kg/m³. Parameters [latex]μ_g and k_g[/latex] are obtained:

[latex]\mu_g=\frac{2m}{ ho\overline{C}as}=\frac{2 imes 2300}{0.9 imes 1.66 imes 6.3 imes 19.33}=26.54 \quad \quad \quad \quad (9.208) \ \space \ k_g=\frac{0.88\mu_g}{5.3+\mu_g}=\frac{0.88 imes 26.54}{5.3+26.54}=0.733 \quad \quad \quad \quad (9.207) [/latex]

When the gust velocity is ±50 ft/s (i.e., ±15.25 m/s), the load factor will be

[latex]n=1+\frac{0.733 imes (\pm 15.25) imes V imes 6.3 imes 0.9 imes 19.33}{2 imes 2300 imes 9.81}\Rightarrow n=1\pm 0.02715V\quad \quad \quad \quad (9.206)[/latex]

The cruising speed ([latex]V_C[/latex]) is given as 310 KEAS; therefore,

n = 1 + 0.02715V = 1 + 0.02715 × 310 × 0.5144 = 5.26 (positive value)

n = 1 - 0.02715V = 1 - 0.02715 × 310 × 0.5144 = -3.26 (negative value)

When the aircraft is flying with dive speed ([latex]V_D[/latex]), the gust speed should be ±25 ft/s (i.e., ±7.5 m/s). Hence, the load factor is

[latex]n=1+\frac{0.733 imes (\pm 7.5) imes V imes 6.3 imes 0.9 imes 19.33}{2 imes 2300 imes 9.81}\Rightarrow n=1\pm 0.01315V \quad \quad \quad \quad (9.206)[/latex]

The dive speed ([latex]V_D[/latex]) is given as 480.5 KEAS; therefore,

n = 1 + 0.01315V = 1 + 0.01315 × 480.5 × 0.5144 = 1 + 3.25 = 4.25 (positive value)

n = 1 - 0.01315V = 1 - 0.01315 × 480.5 × 0.5144 = 1 - 3.25 = -2.25 (negative value)

By comparison between the results of sections (a) and (b), we see that the maximum load factor at sea level is higher than the load factor at 10,000 ft (as we expected). Recall that equivalent airspeeds are independent of altitude.
Therefore, we conclude

+[latex]n_{\max}[/latex] = 6.48
-[latex]n_{\max}[/latex]= -4.48

Thus, the coordinates of points D and I are, respectively, (6.48, 310) and (−4.48, 310). The gust V-n diagram is sketched in Figure 9.31.

3. Combined V-n diagram

Now, we have sufficient data to plot the combined V-n diagram. We first place both diagrams (maneuver and gust) in one plot. Then, we identify and mark the intersection points or corner points for each individual diagram (points A, B, D, F, H, I, K in Figure 9.32). Now, we observe that the line BF and the gust line (n = 1 + 0.03436V) have an intersection. This is marked as point C. In addition, the line BF has an intersection with the line between two positive gust lines. This is marked as point E. The last step is to connect all outer intersection points with straight lines; the combined V-n diagram is created. As a check for sanity, we make sure that no intersection point or corner point is outside the envelope. Also, no point is in the stall regions.
Figure 9.32 demonstrates the final V-n diagram that includes the gust effect. As a conclusion, we observe that the FAR requires that the maximum allowable load factor should be 6. However, due to the gust, this parameter is increased to 6.48.

(3.9): [latex]AR=\frac{b}{C}=\frac{bb}{Cb}=\frac{b^2}{S}[/latex]

(2.49): [latex]V_s=\sqrt{\frac{2W}{ ho SC_{L_{\max}}}}[/latex]

(9.200): [latex]V_D\geq 1.55V_C[/latex] (acrobatic aircraft)

(9.195): [latex]n=\frac{V^2 ho SC_{L_{\max}}}{2mg}[/latex]

(9.206): [latex]n=1+\frac{k_gV_{gE}V_Ea ho S}{2W}[/latex] where
W denotes the aircraft weight (in N)
ρ is the air density (in kg/m³)
S is the wing area (in m²)
[latex]V_E[/latex] is the aircraft equivalent airspeed (in m/s)
[latex]V_{gE}[/latex]is the gust equivalent speed (in m/s)
a is wing lift curve slope during the gust encounter (in 1/rad)

(9.208): [latex]\mu_g=\frac{2m}{ ho Cas}[/latex] where
C is the wing mean geometric chord (in m)
m denotes the aircraft mass (in kg)

Table 9.9 Gus speed for constructing gust-induced load

No.	Aircraft type	Airspeed	Sea level up to 20,000 ft	20,000–50,000 ft
1.	GA-Normal, utility, acrobatic	Cruise speed	50 ft/s	50 ft/s decreases linearly to 25 ft/s
1.	GA-Normal, utility, acrobatic	Dive speed	25 ft/s	25 ft/s decreases linearly to 12.5 ft/s
2.	Commuter	Design speed for maximum gust intensity	66 ft/s	66 ft/s decreases linearly to 38 ft/s
No.	Aircraft type	Airspeed	Sea level up to 30,000 ft	30,000–80,000 f
3.	Transport	Cruise speed	85 ft/s	85 ft/s decreases linearly to 30 ft/s
3.	Transport	Design speed for maximum gust intensity	112 ft/s	112 ft/s decreases linearly to 40 ft/s

Question 9.15: Plot the flight envelope (combined V-n diagram) for the foll...

Related Answered Questions

Verified Answer:

Verified Answer:

Consider an acrobatic aircraft with a mass of 1500 kg diving with a velocity of 100 knot. a. If the aircraft pulls out of the dive with a radius of 150 m, determine the load factor. b. If the maximum allowable load factor is 6, determine the minimum radius for a pull-out of this dive. ...

Verified Answer:

Consider a small remote-controlled (RC) aircraft with the following characteristics: m = 700 g, S = 0.2 m², b = 1 m; e = 0.8; ηP = 0.7, CDo = 0.03, CLmax = 1.2 The airplane is employing a prop-driven electric motor where three cells of 2100 mAh, 12 V Li-Po (lithium polymer) batteries provide ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Consider the single-engine turboprop utility aircraft Cessna 208 Caravan with the following features: mTO = 3995 kg, S = 25.9 m², P = 647 kW, b = 15.88, Vs = 61 knot, Vmax = 185 knot. Assume: CDo = 0.038, ηP = 0.85, e = 0.9. 1. Determine the corner speed of the aircraft. Then, determine the ...

Verified Answer:

Verified Answer:

Verified Answer: